2014西安网络赛1008||hdu5014 二进制

http://acm.hdu.edu.cn/showproblem.php?pid=5014

Problem Description

There is a special number sequence which has n+1 integers. For each number in sequence, we have two rules:

● a_i ∈ [0,n] 
● a_i ≠ a_j( i ≠ j )

For sequence a and sequence b, the integrating degree t is defined as follows(“⊕” denotes exclusive or):

t = (a₀ ⊕ b₀) + (a₁ ⊕ b₁) +···+ (a_n ⊕ b_n)

(sequence B should also satisfy the rules described above)

Now give you a number n and the sequence a. You should calculate the maximum integrating degree t and print the sequence b.

 

Input

There are multiple test cases. Please process till EOF. 

For each case, the first line contains an integer n(1 ≤ n ≤ 10⁵), The second line contains a₀,a₁,a₂,...,a_n.

 

Output

For each case, output two lines.The first line contains the maximum integrating degree t. The second line contains n+1 integers b₀,b₁,b₂,...,b_n. There is exactly one space between b_i and b_i+1(0 ≤ i ≤ n - 1). Don’t ouput any spaces after b_n.

 

Sample Input

42 0 1 4 3

 

Sample Output

201 0 2 3 4

 

解题思路：我们知道一个数有n个二进制位，我们就把它异或成（1<< n）-1,就最大了，因此我们给每一个数从大到小用（1<< n）-1做异或就能得到与之对应的数，注意一定不能从小到大做。

#include <stdio.h>#include <string.h>#include <iostream>using namespace std;typedef __int64 LL;const int N=100055;int a[N],b[N],vis[N];int n;int get(int x){    int num=0;    while(x)    {        x>>=1;        num++;    }    return num;}int main(){    while(~scanf("%d",&n))    {        for(int i=0;i<=n;i++)            scanf("%d",&a[i]);        memset(vis,-1,sizeof(vis));        for(int i=n;i>=0;i--)        {            if(vis[i]!=-1)                continue;            int x=get(i);            int tmp=((1<<x)-1)^i;            b[i]=tmp;            b[tmp]=i;            //printf("%d\n",b[i]);            vis[tmp]=1;            vis[i]=1;        }        LL sum=0;        for(int i=0;i<=n;i++)            sum+=(LL)(i^b[i]);        printf("%I64d\n",sum);        for(int i=0;i<=n;i++)            printf(i!=n?"%d ":"%d\n",b[a[i]]);    }    return 0;}